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2 years ago

Erica brought three and a half yards of fabric if she uses two thirds of it how much will she have left

1 answer:

8 0

3 1/2 - (2/3)(3 1/2) =

7/2 - (2/3)(7/2) = 

7/2 - 7/3 = 

21/6 - 14/6 = 

7/6 or 1 1/6 yds

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A study of 25 graduates of 4-year public colleges revealed the mean amount owed by a student in student loans was $55,051. The s

Harman [31]

Answer:

Step 1

The data represent amount.

A 90% confidence interval for the population mean is,

First, compute t-critical value then find confidence interval.

The t critical value for the 90% confidence interval is,

The sample size is small and two-tailed test. Look in the column headed and the row headed in the t distribution table by using degree of freedom is,

The t critical value for the 90% confidence interval is 1.711.

A 90% confidence interval for the population mean is .

Step 2

It is reasonable to conclude that mean of the population is actually $55000 due to a 90% confidence intrerval for population mean is between $52461.23 and $57640.77 does include $55000.

The data represent amount.

A 90% confidence interval for the population mean is,

First, compute t-critical value then find confidence interval.

The t critical value for the 90% confidence interval is,

The sample size is small and two-tailed test. Look in the column headed and the row headed in the t distribution table by using degree of freedom is,

The t critical value for the 90% confidence interval is 1.711.

A 90% confidence interval for the population mean is .

It is reasonable to conclude that mean of the population is actually $55000 due to a 90% confidence intrerval for population mean is between $52461.23 and $57640.77 does include $55000.

5 0

2 years ago

If a bag contains 12 apples 4 bannans and 8 oranges what is the part whole ratio bannans to fruit

KatRina [158]

Answer:

Step-by-step explanation:

4 to 24 = 1 to 6

4 0

3 years ago

Read 2 more answers

Wich relation is a function of x

faust18 [17]

Answer:

the first one

Step-by-step explanation:

In the other 2 i can see the x repeats itself meaning it isn't a function.

7 0

2 years ago

In a pen with rabbits and chickens someone counted 25 heads and 80 legs a rabbit has four legs and each chicken has 2 legs, how

Komok [63]

I just kind of sat here playing with numbers i dont really know a shortcut but the answer is 10 chickens and 15 rabbits
10*2= 20 each chicken has two legs so thats 10 heads 20 legs
15*4=60 each rabbit has four legs so thats 15 heads 60 legs
15+10=25 heads
20+60=60 legs

6 0

2 years ago

Read 2 more answers

Suppose the horses in a large stable have a mean weight of 975lbs, and a standard deviation of 52lbs. What is the probability th

Lubov Fominskaja [6]

Answer:

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 975, \sigma = 52, n = 31, s = \frac{52}{\sqrt{31}} = 9.34$

What is the probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable?

pvalue of Z when X = 975 + 15 = 990 subtracted by the pvalue of Z when X = 975 - 15 = 960. So

X = 990

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{990 - 975}{9.34}$

$Z = 1.61$

$Z = 1.61$ has a pvalue of 0.9463

X = 960

$Z = \frac{X - \mu}{s}$

$Z = \frac{960 - 975}{9.34}$

$Z = -1.61$

$Z = -1.61$ has a pvalue of 0.0537

0.9463 - 0.0537 = 0.8926

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

7 0

3 years ago